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In-Depth Information
Proof of
Theorem 22.6
(continued from previous page)
Also,
S i () S i ()
. Consequently,
R i () R i ()
. Making this substitution and rear-
ranging terms gives
(22.6)
Δ
R f P
()
+
R f G
()
-
2 R i X
()
From Figure 22.5,
S f () S f () S f ()
+
. Applying Theorem 22.4, we obtain
log
S f ()
+
log S f () 2 log S f () 2
-
, which by the definition of rank, becomes
(22.7)
R f () R f () 2 R f () 2
+
-
Substituting Equation 22.7 into Equation 22.6 yields
(22.8)
Δ
2 R f () 2 R i ()
-
-
2
As for the zig rotation,
, so we can add it to the right side of Equa-
R f () R i () 0
-
>
tion 22.8, factor, and obtain the desired
Δ
3 R f () R i ()
(
-
)
-
2
Finally, we prove the bound for the zig-zig case.
Theorem 22.7
For a zig-zig step,
Δ
3( R f () R i ())
-
-
2.
As before, we have three changes, so the potential change is given by
Proof
Δ
=
R f X
()
-
R i X
()
+
R f P
()
-
R i P
()
+
R f G
()
-
R i G
()
From Figure 22.6,
S f X
()
=
S i G
()
;
their ranks must be equal, so we obtain
Δ
=
-
R
i X
()
+
R f P
()
-
R i P
()
+
R f G
()
We also can obtain
R i P
()
>
R i X
()
and
R f P
()
<
R f X
()
.
Making this substitution and
rearranging gives
(22.9)
Δ
<
R f () R f () 2 R i ()
+
-
(continues next page)
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